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Answer :
i)Given :
Cost of white washing the dome from inside = Rs. 498.96
Also, Cost of white washing 1 \({m}^2\) area = Rs. 2.00
Therefore, CSA of the inner side of dome
= Rs. \(\frac{498.96}{2} {m}^2 \)
= \( 249.48 {m}^2\)
ii) Let the inner radius of the hemispherical dome be r.
Now, CSA of inner side of dome = 249.48 \({m}^2\)
But, CSA = \(2{\pi}{r}^2\)
\(\Rightarrow \) 249.48 \({m}^2\) = \(2 × \frac{22}{7} × {r}^2\)
\(\Rightarrow \) \({r}^2 = (\frac{249.48 × 7}{2 × 22})\) \({m}^2\)
\(\Rightarrow \) r = 6.3 m.
Also, we have,
Volume of air inside the dome = Volume of hemispherical dome
= \(\frac{2}{3}{\pi}{r}^3\)
= \(\frac{2}{3} × \frac{22}{7} × {6.3}^3\) \({m}^3\)
= \(523.908\) \({m}^3\) (approx.)
Therefore, volume of the air inside the dome is \(523.908\) \({m}^3\).